Answer
$${y^2} + 2y + \ln \left| {y - 1} \right| - \frac{1}{2}\ln \left( {{y^2} + 1} \right) - {\tan ^{ - 1}}y + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{2{y^4}}}{{{y^3} - {y^2} + y - 1}}} dy \cr
& {\text{Perform long division}} \cr
& \frac{{2{y^4}}}{{{y^3} - {y^2} + y - 1}} = 2y + 2 + \frac{2}{{{y^3} - {y^2} + y - 1}} \cr
& \int {\left( {2y + 2 + \frac{2}{{{y^3} - {y^2} + y - 1}}} \right)dy} \cr
& {\text{Decompose the integrand }}\frac{2}{{{y^3} - {y^2} + y - 1}}{\text{ into partial fractions}} \cr
& \frac{2}{{{y^3} - {y^2} + y - 1}} = \frac{2}{{{y^2}\left( {y - 1} \right) + y - 1}} = \frac{2}{{\left( {y - 1} \right)\left( {{y^2} + 1} \right)}} \cr
& \frac{2}{{\left( {y - 1} \right)\left( {{y^2} + 1} \right)}} = \frac{A}{{y - 1}} + \frac{{By + C}}{{{y^2} + 1}} \cr
& {\text{Multiply by }}\left( {y - 1} \right)\left( {{y^2} + 1} \right){\text{ and simplify}} \cr
& 2 = A\left( {{y^2} + 1} \right) + \left( {By + C} \right)\left( {y - 1} \right) \cr
& 2 = A{y^2} + A + B{y^2} - By + Cy - C \cr
& 2 = \left( {A{y^2} + B{y^2}} \right) + \left( { - By + Cy} \right) + \left( {A - C} \right) \cr
& {\text{Equate coefficients}} \cr
& A + B = 0 \cr
& - B + C = 0 \cr
& A - C = 2 \cr
& {\text{Solve the system of equations}} \cr
& A = 1,\,\,\,\,B = - 1,\,\,\,C = - 1 \cr
& {\text{Replace the coefficients}} \cr
& \int {\left( {2y + 2 + \frac{2}{{{y^3} - {y^2} + y - 1}}} \right)dy} = \int {\left( {2y + 2} \right)dy + \int {\left( {\frac{1}{{y - 1}} - \frac{{y + 1}}{{{y^2} + 1}}} \right)} } dy \cr
& {\text{Integrate}} \cr
& = \int {\left( {2y + 2} \right)dy + \int {\frac{1}{{y - 1}}} dy} - \int {\frac{y}{{{y^2} + 1}}dy} - \int {\frac{1}{{{y^2} + 1}}} dy \cr
& = {y^2} + 2y + \ln \left| {y - 1} \right| - \frac{1}{2}\ln \left( {{y^2} + 1} \right) - {\tan ^{ - 1}}y + C \cr} $$
